1448. Count Good Nodes in Binary Tree

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1448. Count Good Nodes in Binary Tree

Description

Given a binary tree root, a node X in the tree is named good if in the path from root to X there are no nodes with a value greater than X.

Return the number of good nodes in the binary tree.

Example 1:

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Input: root = [3,1,4,3,null,1,5]
Output: 4
Explanation: Nodes in blue are **good** .
Root Node (3) is always a good node.
Node 4 -> (3,4) is the maximum value in the path starting from the root.
Node 5 -> (3,4,5) is the maximum value in the path
Node 3 -> (3,1,3) is the maximum value in the path.

Example 2:

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Input: root = [3,3,null,4,2]
Output: 3
Explanation: Node 2 -> (3, 3, 2) is not good, because "3" is higher than it.

Example 3:

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Input: root = [1]
Output: 1
Explanation: Root is considered as **good** .

Constraints:

  • The number of nodes in the binary tree is in the range[1, 10^5].
  • Each node’s value is between [-10^4, 10^4].

Hints/Notes

  • N/A

Solution

Language: C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),
 * right(right) {}
 * };
 */
class Solution {
public:
    int res = 0;

    int goodNodes(TreeNode* root) {
        if (!root) {
            return res;
        }
        traverse(root, INT_MIN);
        return res;
    }

    void traverse(TreeNode* root, int min) {
        if (!root) {
            return;
        }
        if (root->val >= min) {
            res++;
        }
        min = max(root->val, min);
        traverse(root->left, min);
        traverse(root->right, min);
    }
};